Introduction to Variational Methods and Finite Elements
Transcript of Introduction to Variational Methods and Finite Elements

Introduction to variational methods and nite elements
1.2.3. Variational formulations of BVP:
Problem: Sove ax = b x =ba
Reformulate the problem:Consider E = 12ax
2 + bxFind x : E(x) = min
xE(x)
ax b
x
x
1. RayleighRitz Method:
Consider a dierential equation
Au = u = f(x) (1a)u(0) = u(1) = (1b)
} Functionalan dimensionvector
Consider the functional: E[u] =10
12(u
)2 + fu dx potential energy functional.
Claim: If u : E[u] = minu
E[u] and u satises (1b) then u solves (1).
Proof:
Let u = u + (arbitrary) (0) = 0 = (1) and C2 so we have a 1 parameter family offunctions which are perturbations of u. The function
Nu
E() = E[u + ] =1
0
12
(u
+
)2+ f(u + ) dx
E(0) =1
0
(u+ ) + f dx
=0
=
10
(u) + f dx
= u10
10
(u
f) dx = 0
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Since is arbitrary we could choose =(u
f)
u f = 0 Euler Lagrange equation for E[u].
How is this useful? Let us assume that U(x) =N1k=1
kk(x)
basis functions
Then
E() =
10
12
[N1k=1
kk(x)
]2+ f(x)
Nk=1
kk(x) dx
For a min:
0 =E
j=
10
(N1k=1
kk
)j + f(x)j(x) dx
=N1k=1
k
10
kj dx+
10
f(x)j(x) dx, j = 1, . . . , N 1
A = b where Akj =1
0
kj dx bj =
10
f(x)j(x)dx.
Example : u = x2u(0) = 0 = u(1)
}u(x) =
x
12(1 x3)
Assume {j} ={1, x, x2, x3
}
U3(x) = c0 + c1x+ c2x2 + c3x3
U3(0) = c0 = 0 u3(1) = c1 + c2 + c3 = 0U3(x) = ax(1 x) + bx2(1 x)
= a{x x2}1
+ b {x2 x3}2
c2x+ c2x2 = c3x+ c3x3c2x(1 x)c2 = c1 c3
A11 =
10
(1 2x)2 dx =1
0
1 4x+ 4x2 dx = x 2x2 + 43x3
1
0
=13
A22 =
10
(2x 3x2)2dx =1
0
4x2 12x3 + 9x4 dx = 43x3 3x4 + 9
5x5
1
0
=20 45 + 27
15=
215
A12 = A21 =
10
(1 2x)(2x 3x2) dx = 16
b1 = +
10
x2(x x2) dx = 120
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b2 =
10
x2(x2 x3)dx = 130
13 1616
215
a
b
=
120
130
a = 1
15b =
16
U3(x) =x
30(1 x)(2 + 5x)
Exact:
uex(x) =x
12(1 x3)
Natural and Essential BC
Notice that the basis functions {i} were required to satisfy the BC u(0) = 0 = u(1). BC that haveto be forced onto the trial solution are called essential BC typically these involve the solutionvalues and not the derivatives. In the case of derivative BC it is possible to build the BC into theenergy functional to be minimized.
Consider{
u(0) = u(1) = u = f(x)
Old energy functional Boundary Condition
E[u] =
10
12(u)2 + fu dx u(1)
E[u] =
10
uu + fu dx u(1)
= uu10
10
(u f) u dx u(1)
= {u(1) }u(1)1
0
(u f)u dx = 0
u = f and u(1) = but says nothing about u(0)!
Dont have to enforce BC at this endpoint.
Eg. 2:
u = x = f uexact = 2 + 7x2 x3
6u(0) = 2 u(1) = 3
essential natural
Let: U(x) = 2 + x(a1 + a2x) = 2 + a1x+ a2x2
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satises homogeneous version of essential BC
UN (x) = +N
i=1
aii(x) =N
i=0
aii(x) a0 = 1 0(x) = 2
E(a) =
10
12
(aii
)2+ f
(aii
)dx 3
(aii(1)
)
0 =E
aj=
10
(aii
)j + fjdx 3j(1)
0 =(i,
j
)ai + (f,j) 3j(1)
0 = Aa+ b
A11 =
10
dx = 1 A21 = A12 =
10
2x dx = 1 A22 =
10
4x2dx =43x3
1
0
=43
b1 =10
(x) x dx 3 = 313
b2 =10
(x)x2 dx 3 = 314[
1 11 4/3
] [a1a2
]=[
10/313/4
] a1 = 4312 a2 =
14
U2(x) = 2 +4312
x x2
4
x 0 0.2 0.4 0.6 0.8 1RR 2 2.707 3.393 4.060 4.707 5.333EX 2 2.699 3.389 4.064 4.715 5.333 Built in
2. General calculus of variations:
I[u] =
ba
F (x, y, y) dx y(a) = 0, y(b) = 0
0 = I =
ba
F
yy +
F
y(y) dx
= yF
y
b
a
b
a
[d
dx
(F
y
) F
y
]y dx
ddx
(F
y
) F
y= 0 EulerLagrange Eq.
Eg.1 (pu) + qu = fu(a) = u(b) =
I[u] =
ba
p(u)2 + qu2 2fu dx+ u
F = p(u)2 + qu2 2fuF
u= 2qu 2f
F
u= 2pu
x
(F
u
) F
u= 2
{(pu) qu + f} = 0
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Eg. 2: SturmLioville Eigenvalue problems:
u(a) = 0 = u(b); +(pu) + qu + ru = 0 (1)
Multiply by v and integrateba+(pu)v + quv dx +
baruv dx = 0
= I[u] =
bap(u)2 qu2 dx
baru2 dx
=I1I2
=I1I2 I1I2
I22=
1I2
(I1 I2)
=
ba2puu 2quu ruu dx
baru2 dx
=2
ba
[(pu) + qu + ru
]u dx
baru2 dx
= 0
= 0 the function u0 which minimizes I[u] is an eigenfunction of (1) and = I[u0] is its eigenvalue.
Higher eigenvalues: n = min(u,k)=0
k=1,...,n1I[u]
constrained minimization problem.3. Method of weighted residuals
What do we do for nonlinear or dissipative problems for which potential energy functionals dontexist or cannot be found easily?
Consider a BVP
Lu = f in = (a, b) (1a)u(a) = ; u(b) + u(b) = (1b)
An approximate solution U wont in general satisfy (1a) and we associate with U the socalledresidual
r(U) = LU fNote that r(u) = 0
exact A whole class of methods are obtained by considering various ways to minimize the residual insome sense, usually:
ba
r(U)i dx = 0
test or weight functions
where U =N
i=1
aii(x) basis function
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1. Collocation: weight functions = (x xi) basis functions polynomials:
Eg:
u u = 0u(0) = 1, u(1) = 0
(A) Let i(x) =(xi+1 x)which satisfy the homogeneous BC for the problem
U(x) = 1 x+ a1(x2 x)+ a2 (x3 x)
r(U) = {a12 + a26x} {a1 (2x 1) + a2
(3x2 1)}+ 1
= (3 2x) a1 a2(3x2 6x 1)+ 1
With x1 = 0 and x2 = 1 as collocation points we have(3 11 4
)(a1a2
)=( 11
) U1(x) = 1 x
(311
)(x2 x) 2
11(x3 x)
If we choose x1 and x2 are chosen to be the zeros of the 2nd degree Legendre polynomial
U2(x) = 1 x 14(x2 x)
(16
)(x3 x)
The exact solution is u(x) =(e ex)(e 1) .
x u(x) U1(x) U1 u U2 U2 u U3 U3 u1/4 0.834704 0.837750 3.05103 0.835938 1.23103 0.769965 6.47 1021/2 0.622459 0.636364 1.39 102 0.625000 2.54 103 0.506144 1.16 1013/4 0.349932 0.360795 0.351563 1.63 103 0.269965 8.00 102
(B) With dierent trial functions:
U(x) = 1 x+ a1 sinx+ a2 sin 3xr(U) = ( sinx+ cosx)a1 3(3 sin 3x+ cos 3x)c2U3(x) = 1 x+ 0.017189 sinx+ 0.011045 sin 3x
using zeros of 2nd degree Legendre polys.
Where to collocate?
What basis fucntions to use?
DeBoor, C. and Swartz, B. , Collocation at Gaussian points, SIAM J. Num. Anal. 10 (1973),582606.
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2. Method of moments: weight functions i = xi.
Eg:
u + u + x = 0u(0) = 0 = u(1)
Let U(x) = a1x(1 x) + a2x2(1 x) + . . .r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1
0
r(U) 1 dx = 0 and1
0
r(U) x dx = 0
116 1112
1122
1920
a1
a2
=
12
13
U(x) = x(1 x)(122649
+110649
x
)
uex(x) =sinxsin 1
x
3. Galerkin method: Expand U(x) and w(x) in terms of the same basis functions.
u(x) =N
i=1
aii w(x) =N
i=1
bii
r
(N
i=1
aii
)(N
i=1
bii
)dx = 0
r
(N
i=1
aii
)j dx = 0
Since the bi are arbitrary.
Eg. u + u+ x = 0u(0) = 0 = u(1)
Let U(x) = a1x(1 x) + a2x2(1 x)r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1
0
r(u) 1(x)dx =1
0
{x+ a1
(2 + x x2)+ a2 (2 6x+ x2 x3)} x(1 x)x2(1 x) dx = 0
310
320
320
13105
a1
a2
=
112
120
a1 = 7369 a2 = 741
U(x) = x(1 x)(
71369
+741
x
)
ue(x) =sinxsin 1
x
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Relationship between Rayleigh Ritz and Galerkin:
Rayleigh Ritz u + u = f = x u(0) = 0 = u(1)
I[u] =
10
u2 u2 + 2fu dx
0 = I =
10
2uu 2uu + 2fu dx
=[uu
]10
10
(u + u f) u dx
0 =1
0
(u + u f)u dx
=
10
r(u)u dx
I(a) =
10
(aii
)(aij
)(
aii)(
ajj
+2f
ajj dx
0 =I
ak= 2
{ai(i,
k)
ai(i,k)
+(f,k)}f = x
RayleighRitz and Galerkin methods are identical for this problem. true in general for linearproblems but not for nonlinear problems.
From WR to the weak form:
WR1
0
(u + u+ x)v dx = 0
u(0) = 0 = u(1) v satisfy homogeneous Dirichlet i.e. v(0) = 0 = v(1)
Say we wanted to express u in terms of functions that are not twice dierentiable then we integrateby parts to throw as many derivatives as needed from u to v.
10
(u + u + x)v dx = uv
1
0
+
10
uv + uv + xv dx = 0.
Now if we let u =
aii and v =
bii we have since v is arbitrary
i
ai(i,j) +
ai(i,j) + (x,j) = 0
Aa = b
where[A]jj = (i,
j) (i,j); b = (x,j)
Identical to RayleighRitz eqs.
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What about a natural BC?
(S){
u + u+ x = 0u(0) = u(1) = .
10
(u + u+ x)v dx = 0 v H10
uv10
10
uv uv xv dx = 0
v(1)1
0
uv uv xv dx = 0
(W)
Find u H1 such that1
0
uv uv xv dx v(1) = 0 v H10
The Finite Element method
F.E. Model Problem
u + f = 0u(a) = G; u(b) = H
}(S)
N i(x) =
Ni(x) =W Residual
ba
(u + f
)v dx = 0 cant plug in
uh(x) =N
i=0uiNi(x)
N i (x) = (x xi1) 2(x xi) + (x xi+1)We must therefore relax the continuity required of the trial solution by going to a weak formulation of the problem.
uv
b
a
+
ba
uv + fv dx = 0
b
a
uv dx = Hv(b) +b
a
fv dx
a(u, v) = Hv(b) + (f, v) ()
(W) Given f L2 and constants G,H ndu {u H1, u(a) = g} = H1g such that for allv H10 = {v H1, v(a) = 0} we have thatbauv dx = Hv(b) +
bafv dx
Hk = {u : u(k) L2} Sobolov space
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Claim: If u is a solution of (S) then usatises (W ). Conversely, provided u is suciently dierentiable then if u satises (W ) it follows that u also satises (S).
(S) (W )( See *)(W ) (S)
ba
uv dx = Hv(b) +b
a
fv dx for all v H10
uv
b
a
b
a
uv dx = hv(b) +b
a
fv dx
ba
(u + f)v dx+ {H u(b)}v(b) = 0
v arbitrary u + f = 0 u(b) = H
natural BC
Using Finite Element Basis Functions: It is useful in the calculation of the integrals to transform each of the subintervals in turn to the same standard interval on which we construct cardinalbasis functions
x0 xi x x3
N2N1
N2()N1()
1 10
xi+1hi+1
N1() = 12(1 )
N2() = 12(1 + )
Na() = 12(1 + a)
a ={ 1 a = 1
1 a = 2
Transformation:
(x) = c1 + c2x (xi) = c1 + c2xi = 1 c2 = 2xi+1 xi =
2hi+1
(xi+1) = c1 + c2xi+1 = 1 c1 = (xi + xi+1)hi+1
(x) =2x (xi + xi+1)
hi+1
Similarly x() =hi+1 + (xi + xi+1)
2
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Notice we can express the transformation in terms of Na():
x() =2
a=1
Na()xa
dx
d=
2a=1
dNad
()xa =(dNad
() =a2
)
=12
2a=1
axa =12(xi + xi+1) = hi+12
d
dx=
2hi+1
Galerkin approximation:
Let uh(x) = GN0(x) +N
i=1
Ni(x)ui
vh(x) =N
i=1
Ni(x)vi arbitrary
a(uh, vh
)= hvh(b) +
(f, vh
)
b
a
(N
i=1
uiNi(x)
) Nj=1
viNj(x)
dx = h vjNj(b) +
ba
fN
j=1
vjNj(x) dx
vj arbitrary N
i=1
ui(N i , N
j
)= hNj(xN ) + (Nj , f) j = 1, . . . , N
Ku = fstiness matrix
Kij =(N i , N
j
)=
ba
N iNj dx
fj = hNj(xN ) + (Nj , f)
i 2 i 1 i i + 1 i + 2
Notice :
(a) Kij = Kji Symmetric
(b) Kij is positive denite
1ijN
uiKijuj =ij
uia(Ni, Nj)uj = a(uh, uh
)=
ba
(u)2
dx 0 pos. semi def.
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not included
(uh)
= 0
= uh = const but since uh(0) = 0
= uh(x) =
uiNi(x) = 0
form a basis ui 0. uTKu 0 = 0 u = 0
(c) Kij = 0 if i j 2
Tridiagonal system  like nite dierences.
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The trick to doing the integrals is to calculate them element by element: so called assembly of thestiness matrix:
e 2 e 1 e e + 1Element stiness matrix
Kij =
ba
N iNj dx =
Ne=1
xexe1
N i(x)Nj(x) dx
=N
e=1
keab
keab =
xexe1
N a(x)Nb(x) dx
=
11
dNad
()(x)dNbd
(x) dxd
d
=2he
a2
b2
11
d =abhe
=(1)a+b
he
[keab] =1he
1 11 1
and keab = 0 if a, b {e 1, e}
Na() =12(1 + a)
N a() =a2
d
dx=
2xe xe1 =
2he
Assembly:
FIGURE assume he = h e i.e. uniform mesh
1h
2 11 2 1
. . .. . .
1 1
u1
...
un
=
F
To calculate the force for a general f(x):
fj =
ba
f(x)Nj(x) dx
We assume f(x) fh(x) =jfjNj(x)
The force vectors are also assembled
fj =e=1
xexe1
f(x)Na(x) dx
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fea =2
b=1
fb
xexe1
Nb(x)Na(x)dx
Simple Examples
01
N1N0
x1 x
(1) N = 1(a)
u = 0 f 0u(0) = G u(1) = G
}ue = G +Hx
k11 =11 1 = 1
f1 = H 1 +1
0
x 0 dx = H
u0 + u1 = f1 u1 = H + G uh=1(x) = GN0(x) + (G + H)N1(x) = G + Hx
(b)
f(x) = pu + p = 0 up = Ax2 2A+ p = 0
ue(x) = + x p2x2
u(0) = G = u(x) = px u(1) = p = H = (p +H)
ue(x) = G +Hx+ p(x x
2
2
)
f1 = H +
10
x p dx = H +p
2
k11u1 = H +p
2+ G
uh(x) = G + Hx+ px2
does not capture exact solution becauseue {set of piecewise linear functions on (0, 1)} .
however uh(0) = G uh(1) = G +H +p
2= ue(1) exact at nodes
(c)
f = qx
ue(x) = G + Hx+ qx
2
(1 x
2
3
)
f1 = H + q
10
x2dx = H +q
3
u1 = H +q
3+ G
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uh(x) = G + Hx+qx
3uh(1) = G + H +
q
3= ue(1) Galerkin exact at nodes
N1 N2
x2 = 1x1
h = 1/2
(2) N = 2 h = 12(a) f 0: u0 = G
1(1/2)
2 11 1
u1
u2
=
HN1(1) +10
N1 0 dx+ 2G
HN2(1) +10
N2 0 dx
4 22 2
u1
u2
=
2G
H
u1 = G +H
2u2 = G + H
uh(x) = GN0 +(G +
H
2
)N1(x) + (G +H)N2(x)
= G + Hx
(b) f = p:
4 22 2
u1
u2
=
2G + p10
N1 dx
H + p10
N2 dx
=
2G + p/2
H + p/4
u1
u2
=
3p/8 +G + H/2
p/2 + G +H
uh(x) = N0G +N1(x) (3p/8 + G +H/2) +N2(x) (P/2 + G +H)
= G + Hx+N1 3p8 +N2p
2
uh(1/2) = G +H
2+
3p8
= uexact(1/2)
uexact(x) = (G+ Hx+ p(x x2/2)
(c)
f = qx
uh = G + Hx+11q48
N1(x) +q
3N2(x)
uh(1/2) = G +H
2+
11q48
= uexact(1/2)
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Using a nite dierence approximation
u + f = 0 f = pu(0) = G u(1) = H
01 2 3
N = 2
n = 1 :U2 2U1 +G
h2= p 1
h[2U1 U2] = hp+ G
h
n = 2 :U3 2U2 + U1
h2= p U3 U1
2h= H U3 = U1 + 2hH
2U2 2U1 2hHh
= hp = U2 U1 = H + h2p
1h
2 11 1
U1
U2
= [ G/h + hp
H + h2p
]h = 1/2
Exactly the same equations as the FEM.
Some convergence results for the F.E. model problem:
We consideru + f = 0u(a) = G u,x(b) = G
}(1)
(A) The Greens function for (1)
Find G(x, x) : G,xx(x, x) + (x x) = 0G(a, x) = 0 = G,x(b, x)
G,x + H(x x) = c1G + x x = c1x+ c2
(i) G,x(b) +H(b x) = c10 + 1 = c1
(ii) G(a) + a x = a+ c20 + 0 = a+ c2
G(x, x) = (x a) x x
x
(x x)dx ={
0 x < x
1 x > x
= H(x x)x
H(x x)dx =
{0 x < x
(x x) x x
x a
a x b
G = x aG = (x a)
Note that G c0 is piecewise linear.
37

Theorem 1: uh(x) the PWL Galerkin approximation is exact at the nodes; i.e., uh(xi) = uex(xi)
Notation: Let vh = {f span {N1, . . . , NN} where Ni H 0}.
Lemma 1: a(h, u uh) = 0 h vh
Proof:
(W) a(u, ) = (f, ) + H(b) H 0 ()u classical or strong solution(S) (W)
(G) a(uh, h) = (f, h) + Hh(b) () h vh =
{h =
ciNij ; Ni H 0
}vh H 0
() a(u, h) = (f, h) +Hh(b) ( )Subtract () from ( ):
a(u uh, h) = 0 h vha(h, u uh) = a(u uh, h) symmetry
Lemma 2: a(G(x x), ) = ((x x), ) H 0Proof:
Gxx + (x x) = 0G(a) = 0 = Gx(b)
0 =
b0
{Gxx + (x x)
} dx = Gx
ba+
ba
Gxx + dx H 0
a(G,) = ((x x), ) H 0
Lemma 3: u(x) uh(x) = a(G(x x) h, u uh) h vh.
Proof: u uh H 0u H1g and uh vh + {gN0} uh(a) = u(a) = g u uh H 0
H 0a(G(x, x
) h, u uh) = a(G, u uh) a(h, u uh)= ((x x), u uh) 0
L(2) L(1)= u(x) uh(x)
Proof of Theorem: u(xi) = uh(xi)
Let x = xi. Then L(3) u(xi) uh(xi) = a(G (x, xi) h, u uh
)but x = xi G(x, xi) =
ciNi(x) PWL basis functions on net {x0, . . . , xN} G vh Gh vh L(1)= u(xi)uh(xi) =0
38

Notes:
1. True no matter how large or small N is.
2. Not true for F.D. equations
3. For this problem it is as if we knew the exact solution and fed this information in at thenodes. Thus the only error in the Galerkin approximation is the interpolation error:
u uh h2
8u = h
2
8f  since u + f = 0.
4. It is not possible to use this technique for all problems as it relied on the fact thatG vh whichis not necessarily true for all problems. For further analysis see Strang and Fix, p. 3951.
f  = maxi,j
supx(xi,xj)
u(x) <
Theorem: (Error in PWL approximation) Let f PC2,(0, 1) then f N
i=0fiNi(x)
18h
2f 
Proof:
xi xi+1
fi+1f(x)
fiNi(x)
fi for any 0 i N
Let w(x) = (x xi)(x xi+1) = x2 (xi + xi+1)x+ xixi+1w = 2x (xi + xi+1) = 0 x =
(xi + xi+1
2
)w
(xi + xi+1
2
)=(xi+1 xi
2
)(xi xi+1
2
)
e(x) = f(x)N
i=0
Ni(x)fi
Claim: For each x [xi, xi+1] x [xi, xi+1] : e(x) = 12f (x)w(x)
Proof:1) x = xi , xi+1 any x suces.2) Choose an arbitrary x = x. Choose such that
(x) = e(x) w(x) = 0
39

then (x) has 3 zeros on [xi, xi+1] andRolle x [xi, xi+1] : (x) = 0
x x
0
0 0 0
0 0 0
= 0
= 0
But
(x) = f (x) 2 = 0 = 1
2f (x)
maxx[xi,xi+1]
e(x) 12f  max
x[xi,xi+1]w(x) 1
8h2f 
Theorem 2: Assume u c1[a, b]. Then there exists at least one point c (a, b) at which uh,x(c) =uexjx (c).
Proof:
MVT c (xi, xi+1) : u(xi+1) = u(xi) + (xi+1 xi)u,x(c) 
uh(xi+1) uh(xi) + (xi+1 xi)u,x(c)
u,x(c) =uh(xi+1) uh(xi)
(xi+1 xi) = uh,x(x)
N i(x) = 1
xi+1 xi Ni+1(x) =
1xi+1 xi u
h,x(x) =
uh(xi+1) uh(xi)(xi+1 xi)
The Barlow points:If we dont know c then for linear elements uh,x at the midpoints are optimally accurate:
Assume u C3(a, b) and expand u about x = [xi, xi+1]:
u(xi+1) = u() + (xi+1 )u() + 12(xi+1 )2u() +
13!(xi+1 )3u(c1)
u(xi) = u() + (xi )u() + 12(xi )2u() +
13!(xi )3u(3)(c2)
Subtract and divide by xi+1 xi = hi+1u(xi+1) u(xi)
xi+1 xi = u() +
12(xi+1 + xi 2)(xi+1 xi)
xi+1 xi u() +O(h2i )
If = xi then uh,x(xi) =uh(xi+1) uh(xi)
xi+1 xi = u(xi) +
12hi+1u
(xi) +O(h2i )
If we choose =xi+1 + xi
2then uh,x(x) = u
(xi+1 + xi
2
)+ O(h2i )
Conclusion:For linear elements the midpoints are superconvergent with respect to derivatives, i.e. most accurate derivatives are calculated there. These are called the BARLOW points. (See Barlow, Int. J.
40

of Num. Meth. Eng., p. 243251, 1976.
(B) Higher order elements:
Lagrange elements:We use the Lagrange polynomials to construct the basis functions:
Lda() =d+1b=1b=a
( b)/ d
b=1b =a
(a b)
Note: Lda(c) = ac just the right properties
d = 1: FIGURE
L11() =( 1)(1 1) =
12(1 ) = N11 ()
L12() =( (1))1 (1) =
12(1 + ) = N12 ()
d = 2: FIGURE
L21() =( 0)(1 0)
( 1)(1 1) =
12( 1) = N21 ()
L22() =( (1))( 1)(0 (1))(0 1) = (1
2) = N22 ()
L23() =( (1))
(1 (1))(1 0) =12(1 + ) = N23 ()
1 0 1
N23 ()
1 0 1
1 0 1
N21 ()
N22 ()
Note:3
k=1
N2k () =122 1
2 + 1 2 + 1
2 +
122 = 1
Can represent a constant function (or rigid body motion) exactly.
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d = 3 :
L31() =116
(1 )(92 1) = N31 ()
L32() =916
(3 1)(2 1) = N32 ()
L33() = 916
(3 + 1)(2 1) = N33 ()
L34() =116
(1 + )(92 1) = N34 ()
1 2 3 4
1 1/3 1/3 1
42

Example:
u + f = 0u(0) = G u(1) = H
uh =N
i=0
uiNi(x)
u = NTu(W) a(uh, vh) = Hvh(b) + (f, vh)
xi1 1/2(xi + xi1) xi
xm
= 2hi (x xm)ddx =
2hi
keab =
xexe1
N a(x)Nb(x) dx
=
11
N a()2h N b()
2h h2d
ke11 =2h
11
{12(2 1)
}2d =
12h
11
42 4 + 1 d = 22h
[433 +
1
0
]=
73h
ke12 =2h
11
12(2 1)(2)d = 2
h
11
22 d = 2h 4
3
3
1
0
= 83h
ke13 =2h
11
12(2 1)1
2(2 + 1)d = +
24h
21
0
42 1d = 1h
[43
3
]10
=13h
ke22 =2h
11
(2)2d = 8h23
3
1
0
=163h
[ke] =13h
7 8 18 16 8
48 8 7
Let f = p : ue(x) = G + Hx + p(x x22
)
0 2
For 1 element h = 1
u0 u1 u2
13
16 88 7
u1
u2
=
83G + 23 p
H + p6 G3
f1 =
xexe1
p N1(x)dx = h2
11
(1 2)d = h[
3
3
]10
= 2h
3p
f2 = H +h
2
11
12( + 2)d = H + p
h
423
3
1
0
= H +h
6p
43

u1
u2
=
7/48 1/6
1/6 1/3
8G + 2pG + 3H + 1/2p
=
7G/6G/6 + 7/24p + H/2 + 1/12p
4G/3 + 1/3pG/3 + H + p/16
=
G + 3/8p + H/2
G + 1/2p + H
N1(x) = 4x(1 x)
N2(x) = x(2x 1)
uh(x) = GN0(x) +(G +H/2 +
38p
)N1(x) + (G + p/2 + H)N2(x)
= G + (H/2N1 (x) +HN2 (x)) +(38pN1 (x) +
p
2N2 (x)
)
= G +H(12(4x 4x2)+ 2x2 x)+ p(3
8(4x 4x2)+ 1
2(2x2 x)
)
= G +Hx+p
2{3x 3x2 + 2x2 x}
= G +Hx+p
2{2x x2}
= G +Hx+ p(x x
2
2
)the exact solution
44